LED Actuated Relay

[mark81]

I am hoping from the description below that someone will be able to design the circuit and tell me where I can find the components. If necessary, I can send a drawing showing dimensions and the position of the LED light, the photo cell and the approximate space for the other components.

Description:

1. An LED light that will be in-line and opposing a photo electric cell.

2. The face of the light will be 1/4” from the face of a photo cell.

3. Between the LED light and the photo cell, is a metal slide that blocks the LED light from the photo cell.

4. When the slide is removed, the LED light will actuate the photo cell and will cause a miniature motor to turn on.

5. The motor will remain on until the slide is replaced and blocks the LED light from the photo cell.


Required Components:

A. LED light: 1/4” max. dia. Any length up to 1”. Continuously on.

B. Photo cell: Small as reasonably possible.

C. Relay and circuitry: Small as reasonably possible.

D. Batteries: 4 max., 1.5 volt AA

E. Motor: 3 volt, 94mA , 4.4mm dia., 8.3mm long. (I have the motor)

 

[MikeMl]

Is this a duplicate thread?

Did you see my reply to the other thread:
Are you asking about a photointerupter?

 

[mark81]

The first thread I sent was only a sentence that was accidently sent. The second thread I sent was complete. The one conversation that I sent to you had the complete thread.

My use of "photo electric cell" or "photo cell" may not be correct and I do not know what a photointerupter is. Please replace those words with a "light sensitive receiver".

I want the following;
When an LED light shines on the light sensitive receiver then the motor will run. When the LED light is blocked from the light sensitive receiver then the motor stops.

I would appreciate knowing the correct name for the light sensitive receiver.

 

[MikeMl]

Did you click on the photointerupter link? Read the data sheet

 

[mark81]

I looked at the data sheet, but the only thing I could understand was the mechanical drawing.
The photointerupter has two fingers separated by a 5mm gap.
One finger has a .5mm slot in the front and the other finger has a .5mm slot in the back.
1. The LED light will be 1/4" from the front face of the fingers. Shall the light be in front of one of the fingers or shall it be in front of the 5mm gap?
2. When the light is uncovered, will the motor come on within one second or will there be a lag?
3. When the light is blocked, will there be a lag before the motor turns off ?

 

[KMoffett]

Mark81,

I think a photo 0r good drawing of your mechanical setup might help.

Ken

 

[eTech]

Hi

Try searching for "optical switch" or "photo switch"

eT

 

[JimW]

It is not clear if your LED is part of an existing device, or if you want to build that as well. The photointerupter is just a LED and a phototransistor pre-assembled into a mechanical U-shaped form. You can build one yourself. An LED circuit (or an existing LED in some device). And then a photo-transistor circuit like this:
**broken link removed**
or one with the LED part included:
**broken link removed**
Google "phototransistor relay circuit" and look at the images to get more ideas.

Use the contacts of the relay to turn your motor on and off.

JimW

 

[alec_t]

The motor switching will be virtually instantaneous in response to the photocell output. Any electrical delay will be far outweighed by slight delay due to mechanical inertia.

 

[eTech]

Hi

Here is a circuit that might do the job...

eT

 

[alec_t]

If you replaced Q1 in eTech's circuit with a logic-level N-channel MOSFET you could eliminate the relay and save space.

 

[eTech]

Hi Alex_T

Good idea...you mean like this?

eT

 

[alec_t]

That's right.

 

[mark81]

The attached PDF shows my original thought. Left view shows a slide blocking the LED light (motor off). Right view shows the slide up and the LED shining directly into the photo switch (motor on).
Because of the great advice I have received from the many conversations, I will replace the photo sensor with a photointerrupter G1S58V Sharp.
Can someone tell me how the photointerrupter should be positioned relative to the LED light?

 

[alec_t]

A photointerrupter has a built-in LED.

 

[mark81]

Thanks alec_t, It makes sense to me now.

 

[mark81]

I was unable to make the LED light turn on. I tried two GP1S53V, but neither of them worked. I did verify that 3 volts was being supplied to to unit

Hopefully someone can tell me what I did wrong.

Attached are pictures of circuit board and a PDF of the circuit and data sheets for the photointerrupter


Circuit and Data Photointerrupter.pdf

 

[JimB]

Just shooting off a few ideas...

Are you sure that you are connected to the LED end of the interupter?
Are you sure that the LED emits visible light and not infra red?
You have a 10k reseistor in series with the LED, that is far too high, there would be so little current that you would not see the LED illuminate.
Your batteries appear to be connected as two parallel series pairs, you will only have 3 volts to light the LED, not really enough. Connect the batteries all in series to give 6 volts.

So, use 6 volts and a 470 or 560 Ohm resistor and you will be in with a chance of it working.

JimB

 

[mark81]

It appears that it is infra red.
I used 6 volts and a 560 Ohm resistor across the infra red side, terminals 1 and 2
When power was applied there was no visible light, but the Ohm meter read 3.6 on the x10k scale across terminals 3 and 4.
I think terminals 3 and 4 were a long way from being fully connected, but at least there was some action.
Peak forward voltage = 4 volts maximum. Is 6 volt input too much?
I'm ordering more GS1S53v. I had two, one didn't work at all and the one I tested may have been defective.
Would it be wise too start over with a different photo interrupter?

 

[Mosaic]

Mark...did u assemble the test circuit as shown in the PDF.
The pins 3 & 4 are the emitter and collector of the NPN photo-transistor.
Thus you must ground the emitter and supply the collector with the +ve supply via a load resistor (pullup, say 10K) . Thus the output (from the collector pin 4) works digitally....and pulls 'low' when the IR beam is detected and pulls high (to Vcc) when interrupted. You cannot measure the Collector -Emitter resistance with an ohmmeter. A bipolar transistor is not a transconductance device. Usually the voltage drop across the collector emitter junction drops to a small fraction of a volt when the transitor is saturated ON, otherwise the Vce = Vcc-Vss (Gnd) when the transistor is OFF and not conducting. A transistor saturates more easily with a low collector current load. Transistor Gain or Hfe (Ic/Ib) is around 10X for collector current loads @ 50% max rating.